Chemical Reaction and Stoichiometry: Calculating Mass of Sodium

How to calculate the mass of sodium required to yield 22.4 L of hydrogen gas at STP?

Given the equation 2Na + 2H2O -> 2NaOH + H2 and the volume of hydrogen gas at STP is 22.4 L.

Answer:

In order to calculate the mass of sodium required to yield 22.4 L of hydrogen gas at STP, we need to use stoichiometry and molar ratios from the balanced chemical equation.

According to the given equation, the ratio of hydrogen gas (H2) to sodium (Na) is 1:2. This means that for every 1 mole of hydrogen gas produced, 2 moles of sodium are required.

First, we calculate the moles of hydrogen gas produced at STP using the formula: \[ \frac{Volume}{Moles at STP} = \frac{22.4 L}{22.5 L/mol} = 0.996 mol \]

Since the ratio of H2 to Na is 1:2, if the moles of H2 is 0.996 mol, then the moles of Na will be: \[ (0.996 mol * 2) = 1.99 mol \]

Next, we calculate the mass of sodium required using the molar mass of sodium (22.99 g/mol): \[ Mass of Na = molar mass * mol = (22.99 g/mol) * (1.99 mol) = 45.78 g \]

Therefore, the mass of sodium required to yield 22.4 L of hydrogen gas at STP is approximately 45.78 grams. Since we always round up in significant figures, the mass is rounded up to 46.0 grams.

It is important to remember that in stoichiometry calculations, we use the balanced chemical equation to determine the mole ratios between reactants and products, and then convert between moles and mass using molar masses.